One can solve Poisson’s problem $-\Delta u = f$ in $d$ dimensions with homogeneous Dirichlet boundary conditions using a mixed formulation as explained below:

Let $\sigma = \nabla u$, then for a sufficiently smooth function $\tau$, by Green’s theorem

\begin{align*}

(\sigma, \tau) &= (\nabla u, \tau) \\

&= -(u, \textrm{div } \tau).

\end{align*}

Again, choosing $v$ a function sufficiently smooth, we have

\begin{align*}

f = -\textrm{div } \sigma \implies (f, v) = (-\textrm{div } \sigma, v).

\end{align*}

This gives the saddle-point problem: find $(u, \sigma) \in V \times M$ such that

\begin{align*}

(\sigma, \tau) + (u, \textrm{div } \tau) &= 0\\

(\textrm{div } \sigma, v) &= -(f, v)

\end{align*}

hold for all $(\tau, v) \in V \times M$. Note that we don’t have to take a derivative of $u$, hence it’s natural to try $M = L^2$, but what about the space $V$?

One very easy choice to guess is $V = [H^1(\Omega)]^d$ as we want the divergence to be all defined, but unfortunately this doesn’t work as the gradient of the solution to Poisson’s problem can easily not be in $[H^1(\Omega)]^d$.

In order to illustrate this, consider $u =\left(r^{2/3}-r^{5/3}\right) \sin \left(\frac{2 \theta }{3}\right)$ on the domain of the unit circle with bottom left quarter taken out. It’s not hard to see that $u = 0$ on the boundary of the domain, and we can easily find the $f$ such that it satisfies Poisson’s equation. Now, we can either calculate the gradient exactly or argue as follows.

First, recall how to take a gradient in polar coordinates. Note that $\partial_r u \approx r^{-1/3}$ plus higher order terms and that $\frac{1}{r}\partial_\theta u \approx r^{-1/3}$ plus higher order terms also. Now, one can easily calculate the $H^1$ seminorm to see that the derivative is unbounded as we’re integrating over $[0,1]$ with $(r^{-4/3})^2r$ terms (the extra $r$ comes from the change of basis from polar integration).

The above is an example of why the space $H(\textrm{div})$ is needed.